12-10
Application Note
Using a Shunt Resistor to Dissipate
Energy
Rev. 2
Jan. 2011
2
Scope
When a moving motor and load are required to decelerate rapidly, the kinetic
energy of the load drives the motor's shaft to generate energy back to the servo
drive. This is called “regenerative energy”.
In most cases, regenerative energy is stored by the motor's internal resistance,
servo drive and power supply, but in some cases, when high speed and inertia are
involved, the energy becomes too great and causes the voltage on the bus (Vbus) to
rise. A shunt resistor and a control provide a simple and effective solution to
protect the power supply and servo drive by sinking energy though the shunt
resistor.
This application note describes methods to calculate the resistance and power of
the shunt resistor.
System Diagram
The shunt resistor is placed between the VP+ and PR terminals of the servo drive
using a special control circuit, as shown below:
When the control circuit detects the presence of a voltage rise above a given
threshold level, it connects the shunt resistor, and the energy sinks from the VP+
terminal to the PR terminals and dissipates as heat.
The resistor should have a low enough resistance to enable it to absorb the
regenerative energy. It should also have a sufficient power rating.
Shunt Resistor Value Calculations
For a given application, the maximum resistance of the shunt resistor is derived
from the motor's characteristics, the value of Vbus, the braking torque and the
maximum speed.
The mechanical power to be dissipated is given by the following formula:
P = T ⋅ω
where:
T is the torque in Nm,
ω is the angular velocity in rad/sec.
3
The braking torque can be derived from the motor's torque constant (Kt) and by
measuring the value of the current on deceleration.
The mechanical power is dissipated on the motor itself and the shunt resistor. It is
also partially absorbed by the capacitance on the bus.
For the purposes of this Application Note, we will neglect the bus capacitance
absorption and concentrate on the shunt resistance.
The power dissipation on the motor itself is actually the copper loss and can be
calculated from the formula:
Pmot = I Q2 Rpp
where:
R pp is the phase-to-phase resistance of the motor,
I Q is the current at brake time.
The additional power Pshunt = P − Pmot needs to be dissipated on the shunt resistor.
Otherwise, the bus voltage Vbus will rise and may cause damage to the bus
components that cannot withstand the high voltage level.
Normally, we allow a voltage rise up to 90% of the drive's maximum voltage rating
before engaging the shunt resistor (unless there are other voltage-sensitive
components on the bus). When the shunt resistor is engaged, current flows
through the resistor, causing Pshunt to be dissipated as heat. For good practice, we
recommend using a maximum of 70% pulse-width modulation (PWM) to the Rshunt
switching transistor.
The shunt resistance is therefore calculated at the maximum allowable voltage
from the formula
Rshunt =
(0.7 × Vmax )2
Pshunt
2
≅
Vmax
2 × Pshunt
where:
Vmax is the maximum level of Vbus (the voltage level at which the shunt resistor is
engaged).
Shunt Resistor Power Calculations
The shunt resistor should dissipate a maximum peak power when the motor is at
its maximum speed ωmax and its maximum deceleration torque T:
Pmax = T ⋅ ω max − I Q Rpp
2
Assuming that the deceleration torque is constant, we may assume that the
current I Q is also constant. Since the speed is constantly decreasing, the power
that needs to be dissipated is also decreasing.
4
At a certain speed ωi , the shunt resistor's power will be zero:
ωi =
I Q Rpp
Kt
The power on the shunt resistor actually dissipates within a very short period of
time.
Below is a typical motion diagram that shows the power dissipation on the shunt
resistor.
The resistor's average power is derived from the value of Pmax calculated above
and from Δ t , which is influenced by the value of the deceleration a .
For a given deceleration a and Pmax , the shunt resistor's average power is:
Pshunt =
(ωi − ω max )⋅ Pmax
2⋅a
5
Example
Regeneration current
M
Shunt drive pulses
In this example, we assume the following motor data:
V shunt =
0.9 x 200 V = 180 V
Speed =
2000 rpm
Braking current =
5A
R=
10 Ω
Ke =
40 V/krpm
Therefore,
K e [V/krpm] =
40 V
= 0.3820 V sec/rad
(1000 ⋅ 2 ⋅ π / 60)
Thus K t = 0.3820 Nm/A
The dissipated mechanical power is:
⎛ 2000 ⋅ 2 ⋅ π ⎞
P = Tω = K t Iω = (0.3820 ⋅ 5) ⋅ ⎜
⎟ = 400 W
60
⎠
⎝
The copper loss on the motor is:
Pmot = I Q2 Rpp = 52 ⋅ 10 = 250 W
This leaves 400 W – 250 W = 150 W to be dissipated by the shunt resistor.
The maximal shunt resistance is:
Rshunt =
2
1802
Vmax
=
= 108 Ω
2 ⋅ Pshunt 2 ⋅150 W
6
Therefore, the final resistance is Rshunt ≤ 108 Ω .
Now, let us assume the following deceleration: a = −1000 rpm/sec
2
Then we have:
(−1000 ⋅ 2 ⋅ π / 60)
ω
= −104.72 2
2
sec
sec
5 A ⋅10 Ω
ωi =
= 130.89 rad/sec
0.3820
a=
Therefore the power of the shunt resistor is:
Pshunt
⎛
⎛ 2 ⋅π
⎜130.89 − 2000⎜
⎝ 60
=a⎝
2 ⋅ (−104.72)
Pshunt ≥ 56 W
⎞⎞
⎟ ⎟ ⋅150
⎠⎠
= 56 W
7
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